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I have a hard time coming up with a reason to why this is the case, anyways my task formulated as the book describes is this:

Suppose $H$ is a subspace of $\mathbb{R}^n$ and suppose that $K$ is another subspace of $\mathbb{R}^n$ such that $K\subset H$.

Show that $$\dim(K) < \dim(H)$$

Now i know this is very similar to How can a subspace have a lower dimension than its parent space?, but then again not completely. So I ask you dear users of this forum to help a fellow out.

The only thoughts that come to my mind is if i somehow knew the span of $K$ and $H$ then it would be almost trivial (in theory) to take the reduced row echelon form and determine by the pivot columns the number by dimensions.

zhengchl
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Thomas
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2 Answers2

9

If $K \subsetneq H$ then there is a vector $h \in H$ that isn't in $K$. Choose a basis $(k_1,\ldots,k_l)$ for $K$. I claim that the vectors $k_1,\ldots,k_l,h$ are linearly independent. Otherwise, $h$ would be a linear combination of the $k_i$ (by the linear independence of the $k_i$) and thus belong to $\operatorname{span}(k_1,\ldots,k_l) = K$. Thus $H$ contains strictly more linearly independent vectors than $K$, which implies that $\dim K < \dim H$.

Alex Provost
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Is there anything there that says "proper" subspace? Any vector space is a subspace of itself so unless you are saying "proper subspace" this statement simply isn't true.

user247327
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